Four grams of hydrocarbon (C_xH_y) on complete combustion gave 12grams of CO₂. What is the empirical formula of the hydrocarbon ? (C = 12 ; H = 1)

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CH₃

C₄H₉

C₃H₈

answer is D.

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Detailed Solution

Weight of C_xH_y = 4 gm

Weight of CO₂ = 12 gm

$%;of;Carbon;=;large frac{12}{44} times frac{W_{CO_2}}{W_{OC}} times 100$

$large %;of;Carbon;=;frac{12}{44} times frac{12}{4} times 100$

= 81.8%

% of Hydrogen = 18.2%

$large begin{array}{*{20}{c}} {}&{C:H}\ {Mass;ratio}&{81.8:18.2}\ {Gram;atom;ratio}&{frac{{81.8}}{{12}}:frac{{18.2}}{1}}\ {Gram;atom;ratio}&{6.82:18.2}\ begin{array}{l} Gram;atom;ratio\ Gram;atom;ratio\ Gram;atom;ratio\ Gram;atom;ratio end{array}&begin{array}{l} frac{{6.82}}{{6.82}}:frac{{18.2}}{{6.82}}\ 1:2.67\ 1(3):2.67(3)\ 3:8 end{array} end{array}$